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Chapter 4: Problem 70

Use limit methods to determine which of the two given functions grows faster, or state that they have comparable growth rates. $$x^{2} \ln x ; \ln ^{2} x$$

Short Answer

Expert verified
Answer: The function \(x^2 \ln x\) grows faster than the function \(\ln^2 x\) as \(x\) approaches infinity.

Step by step solution

01

Define the limit equation

First, we define a limit equation that will help us understand the growth rate comparison between the given functions. Consider the limit: $$\lim _{x \rightarrow \infty} \frac{x^{2} \ln x}{\ln ^{2} x}$$
02

Apply L'H么pital's Rule

Since the limit has the form \(\frac{\infty}{\infty}\), we can apply L'H么pital's Rule. To do this, we will differentiate both the numerator and denominator concerning \(x\): $$\lim _{x \rightarrow \infty} \frac{(2x \ln x + x^2 \frac{1}{x})}{(2 \ln x \frac{1}{x})} = \lim _{x \rightarrow \infty} \frac{2x \ln x + x}{2 \ln x \cdot \frac{1}{x}}$$
03

Simplify the limit equation

Let's simplify the limit equation by canceling out any common factors and rearranging terms: $$\lim_{x \rightarrow \infty} \frac{2x^2 \ln x + x^2}{2 \ln x} = \lim_{x \rightarrow \infty} \frac{x^2(2 \ln x + 1)}{(2 \ln x)}$$
04

Apply L'H么pital's Rule again

The limit still has the form \(\frac{\infty}{\infty}\), so we can apply L'H么pital's Rule again. We differentiate both the numerator and denominator concerning \(x\): $$\lim_{x \rightarrow \infty} \frac{d}{dx}[x^2(2 \ln x + 1)]{d/dx(2 \ln x)} = \lim_{x \rightarrow \infty} \frac{2(x^2(2 \frac{1}{x} + \frac{2}{x})) + 2x(2 \ln x + 1)}{2 \cdot \frac{1}{x}}$$
05

Simplify the limit equation again

Let's simplify the new limit equation: $$\lim_{x \rightarrow \infty} \frac{2x^2 + 4x \ln x + 2x}{2 \cdot \frac{1}{x}} = \lim_{x \rightarrow \infty} x \cdot \frac{2x + 4 \ln x + 2}{2}$$
06

Evaluate the limit

Now, let's evaluate the limit as \(x\) approaches infinity: $$\lim_{x \rightarrow \infty} x \cdot \frac{2x + 4 \ln x + 2}{2} = \infty$$ Since the limit is infinite, we conclude that the first function, \(x^2 \ln x\), grows faster than the second function, \(\ln^2 x\).

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